Help with CO2 soulubility equilibrium (1 Viewer)

[Dashin_Dan]

I have very little notes on the CO2 solubility equilibrium. Can anyone help me with this? Any would be great.

 

[nomz]

try the csu hsc online site-
http://hsc.csu.edu.au/chemistry/core/acidic
there's a few notes there...

 

[funking_you]

I have very little notes on the CO2 solubility equilibrium. Can anyone help me with this? Any would be great.

Normally i wouldn't do this, however i've attached a worksheet on Equilibrium Systems, it is very simple to understand.
Equilibrium is such an important concept, and once you get the hang of how to apply its concepts, you'll actually enjoy this part of the syllubus.

Read through, and work through the questions, and i guarantee you'll master this concept.

Any more questions, email me at
george@chemistrycoach.com.au

Cheers,
George


Get the worksheet.....

 

[richz]

thnx george

 

[funking_you]

Glad be of assistance! if you require any help with the exercises in the worksheet just post your questions here

 

[lfc_reds2003]

Glad be of assistance! if you require any help with the exercises in the worksheet just post your questions here

ill take u up on that...

H2(g) + I2(g) = 2HI(g) Pressure increased

which side is favoured (aren't there same ratio of moles on each side??)

 

[Danni07]

I asked my teacher he said it's almost like a trick question, and the equilibrium doesn't shift

 

[Trev]

aural_sax05;
since the same ratio of moles on both sides, the equilibrium doesn't shift to either side.
correct me if i'm wrong chemcoach, ps. thanx for sheet!

 

[funking_you]

Download the EQUILIBRIUM SYSTEM'S document in my early thread.

Pressure and Volume changes only affect GASEOUS equilibrium systems, and if there is an EQUAL amount of gases moles on each side of the reaction, the changes (increase/decreases) have no effect on the equilibrium.

Why?
This is because pressure is directly related to the number of molecules, and hence if there is the same number of molecules (and hence moles) on both sides, the pressure would be just about the same too.

Its not a 'TRICK' question, just examines whether you understand the finer point of equilibriums.

TREV, you got it correct, well done mate!

Keep studying everyone,
George

 

[lfc_reds2003]

good i just leanrt that today

thanks anyway y'all

Gl in chemistry all u 05ers

 

[song]

the solubility of carbon dioxide in water depends on:
- pressure
- temperature

The equilibrium of carbon dioxide in water can be described using the following:
CO2 (g) <=====> CO2 (aq)
CO2(aq)+ H2O(l) <======> H2CO3 (aq)
H2CO3(aq) + H2O(l) <======> HCO3-(aq) + H3O+(aq)
HCO3-(aq) + H2O(l) <=====> H3O (aq)+ + CO3 2- (aq)

now, increasing the pressure means that the equilibrium will shift to the right (as described by Le chat's priniciple) because the sytem wants to decrease the CO2 (g) by dissolving the gas and therefore reducing the pressure of the system.

now, increasing the temperature means that the system favours the endothermic reaction.

 

[xiao1985]

on acidity of the water as well,

as the carbonic acid could react with base...